First of all, I've tried searching for the derivation but I was unsuccesful. I'm trying to prove the following relation
$\left(\sigma_{\mu\nu}F^{\mu\nu}\right)^{2}=2F^{\alpha\beta}F_{\alpha\beta}+i\gamma^{5}\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}, $
which was used by Schwinger (1951) to obtain a QED effective lagrangian. However, I'm stuck with my derivation. Here's the attempt
$\begin{aligned}\left(\sigma_{\mu\nu}F^{\mu\nu}\right)^{2} & =\sigma_{\mu\nu}\sigma_{\alpha\beta}F^{\mu\nu}F^{\alpha\beta}\\ & =1/2\left\{ \sigma_{\mu\nu},\sigma_{\alpha\beta}\right\} F^{\mu\nu}F^{\alpha\beta}+1/2\left[\sigma_{\mu\nu},\sigma_{\alpha\beta}\right]F^{\mu\nu}F^{\alpha\beta} \end{aligned} $
which can be separated in the first and second terms. Because
$1/2\left\{ \sigma^{\mu\nu},\sigma^{\alpha\beta}\right\} = \left(\eta^{\mu\alpha}\eta^{\nu\beta}-\eta^{\nu\alpha}\eta^{\mu\beta}+i\gamma^{5}\epsilon^{\mu\nu\alpha\beta}\right),$
it is easy to see that the first term is the complete answer, by applying the property that the electromagnetic tensor is antisymmetric $(F^{\beta\alpha}=-F^{\alpha\beta})$:
$1/2\left\{ \sigma_{\mu\nu},\sigma_{\alpha\beta}\right\} F^{\mu\nu}F^{\alpha\beta} = 2F^{\alpha\beta}F_{\alpha\beta}+i\gamma^{5}\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta}$.
But I can't seem to actually prove that the second term is nule
$\begin{aligned}1/2\left[\sigma_{\mu\nu},\sigma_{\alpha\beta}\right]F^{\mu\nu}F^{\alpha\beta} & =i\left(\eta^{\nu\alpha}\sigma^{\mu\beta}-\eta^{\mu\alpha}\sigma^{\nu\beta}-\eta^{\nu\beta}\sigma^{\mu\alpha}+\eta^{\mu\beta}\sigma^{\nu\alpha}\right)F_{\mu\nu}F_{\alpha\beta}\\ & i\left(\sigma^{\mu\beta}F_{\mu}^{\,\alpha}F_{\alpha\beta}-\sigma^{\nu\beta}F_{\,\nu}^{\alpha}F_{\alpha\beta}-\sigma^{\mu\alpha}F_{\mu}^{\,\beta}F_{\alpha\beta}+\sigma^{\nu\alpha}F_{\,\nu}^{\beta}F_{\alpha\beta}\right) \end{aligned} $.
If anyone has a derivation or a clue on how to proceed it is highly appreciated. Thanks in advance.
Exchanging $\mu\nu$ with $\alpha\beta$ changes the sign of the anticommutator while preserving $F^{\mu\nu}F^{\alpha\beta}$, thus multiplying the term by $-1$. But such an exchange preserves everything else in the problem. Another way to put it is to use $A^2=\frac12\{A,\,A\}$ to obtain$$(\sigma_{\mu\nu}F^{\mu\nu})^2=\frac12\{\sigma_{\mu\nu}F^{\mu\nu},\,\sigma_{\alpha\beta}F^{\alpha\beta}\}=\frac12\{\sigma_{\mu\nu},\,\sigma_{\alpha\beta}\}F^{\mu\nu}F^{\alpha\beta}.$$